Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0^{+}}(\sin x)^{x} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{e} \).

Step 1: Identify the indeterminate form

To find \( \lim _{x \rightarrow 0^{+}}(\sin x)^{x} \), start by recognizing any indeterminate forms when \( x \to 0^{+} \). Notice that \( \sin x \to 0^{+} \) as \( x \to 0^{+} \) and \( x \to 0 \) from the positive side. So the expression \((0^{+})^{0}\) indicates a \(0^0\) form, which is an indeterminate form suitable for l'Hôpital's Rule.

Step 2: Transform the problem using logarithms

To simplify the expression, take the natural logarithm of both sides: \( y = (\sin x)^{x} \), then \( \ln y = x \ln(\sin x) \). So, the problem is transformed into finding the limit of \( \ln y \). Thus, \( \lim_{x \to 0^{+}} \ln y = \lim_{x \to 0^{+}} x \ln(\sin x) \).

Step 3: Apply l'Hôpital's Rule

The expression \( x \ln(\sin x) \) can be analyzed as \( \frac{\ln(\sin x)}{1/x} \) which is in the form \( \frac{\infty}{\infty} \) for \( x \to 0^{+} \), allowing us to use l'Hôpital's Rule. Find the derivatives: the derivative of \( \ln(\sin x) \) is \( \frac{\cos x}{\sin x} = \cot x \), and the derivative of \( 1/x \) is \( -1/x^2 \). Applying l'Hôpital's Rule, \[ \lim_{x \to 0^{+}} \frac{\ln(\sin x)}{1/x} = \lim_{x \to 0^{+}} \frac{\cot x}{-1/x^2}. \] When simplified, \( \lim_{x \to 0^{+}} (-x^2 \cot x) \).

Step 4: Simplify and find the limit

Now evaluate \( \lim_{x \to 0^{+}} (-x^2 \cot x) \). Since \( \cot x = \frac{\cos x}{\sin x} \), this becomes:\( \lim_{x \to 0^{+}} \left(-x^2 \cdot \frac{\cos x}{\sin x} \right)\) = \( \lim_{x \to 0^{+}} -x^2 \cdot \frac{1}{x} = \lim_{x \to 0^{+}} -x * \frac{1}{\sin x}\). Given \(\sin x \approx x\) as \(x \rightarrow 0\), therefore this tends to \(-1\).

Step 5: Exponentiate the result to obtain the limit of the original function

Since we have found \( \ln y = \lim_{x \to 0^{+}} x \ln(\sin x) = -1 \), we exponentiate to return to the original variable: \( \lim_{x \to 0^{+}} y = e^{-1} = \frac{1}{e} \).

Key Concepts

Indeterminate Form

Indeterminate forms are expressions in calculus that do not have a well-defined limit or value. They arise when the limits involve basic arithmetic operations that initially seem undefined. Common types include the forms \(0/0\), \(\infty/\infty\), \(0^0\), \(1^\infty\), and others. In the context of limits, these forms indicate that further analysis is needed to determine the actual limit.

• For instance, \((0^+)^0\) is an indeterminate form because the base, as it approaches zero, and the exponent, lead to a scenario that isn't straightforwardly resolved as either zero, one, or any other concrete number without deeper investigation.
• L'Hôpital's Rule is often used when dealing with an indeterminate form, allowing us to differentiate the numerator and the denominator of a fraction separately to find a clearer path to the limit.

Recognizing these forms is crucial since they help identify when advanced calculus methods, like l'Hôpital's Rule, are necessary.

Natural Logarithm

Natural logarithms are logarithms with the base \(e\), an irrational constant approximately equal to 2.71828. The notation \(\ln\) is commonly used to denote the natural logarithm. Natural logarithms are particularly useful in calculus, especially when involving rates of growth or decay, because they provide a way to transform power expressions into linear equations, simplifying the process of differentiation or integration.

• For example, in this exercise, the expression \((\sin x)^x\) is transformed into \(x \ln(\sin x)\) by taking the natural logarithm of both sides. This transformation was key for applying l'Hôpital's Rule.
• Taking logarithms can convert multiplication into addition, powers into multiplication, and thus simplify complex expressions for calculus operations.

Understanding how and when to use natural logarithms is vital in simplifying functions and solving limits where indeterminate forms like \(0^0\) are present.

Limit of a Function

The limit of a function describes the behavior of the function as its argument approaches a particular point. Essentially, it helps in finding what value the function approaches as the input nears a specified point.

• In this context, we are interested in \(\lim_{x \to 0^+}(\sin x)^x\), focusing on how the function behaves as \(x\) approaches 0 from the positive side.
• Limits are fundamental in calculus as they form the basis for discussing continuity, derivatives, and integrals; they represent a precise way to talk about approaching values rather than just the values at certain points.

Understanding limits includes recognizing when a specific technique like l'Hôpital's Rule is needed due to the presence of an indeterminate form.

Calculus Techniques

Calculus offers various techniques for dealing with limits and determining the behavior of functions, particularly when direct substitution in the function's formula yields an indeterminate form. One such powerful technique is l'Hôpital's Rule, which applies to ratios of functions that yield indeterminate forms like \(0/0\).

• To use l'Hôpital's Rule, one differentiates the numerator and the denominator separately until a determinate form is achieved, allowing for a conclusive evaluation of the limit.
• In our exercise, applying l'Hôpital's Rule to the expression \(x \ln(\sin x)\) involved converting this into a ratio suitable for differentiation and simplification through the rule.

This approach highlights the usefulness of derivatives in uncovering limits previously obscured by indeterminate forms. With these techniques, calculus becomes a versatile tool for solving complex problems involving limits and functions.